Two slits are made 0.1 mm apart, and the screen is placed 2 m away. The fringe separation when a light of wavelength 500 nm is used is:
1 cm
Understanding Fringe Separation in Double-Slit Interference
This problem asks us to calculate the fringe separation in a Young's double-slit experiment given the slit separation, the distance to the screen, and the wavelength of the light used.
Given Parameters for Double-Slit Experiment
We are provided with the following information:
- Slit separation, \(d = 0.1 \text{ mm}\)
- Distance from slits to screen, \(D = 2 \text{ m}\)
- Wavelength of light, \(\lambda = 500 \text{ nm}\)
Converting Units
Before using the formula, it's essential to convert all given units to the standard SI unit, meters (m).
- Slit separation, \(d\): \(0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 1 \times 10^{-4} \text{ m}\)
- Distance to screen, \(D\): \(2 \text{ m}\) (already in meters)
- Wavelength, \(\lambda\): \(500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}\)
Formula for Fringe Separation
In a Young's double-slit experiment, the fringe separation (\(\beta\)), also known as fringe width, is given by the formula:
\[\beta = \frac{\lambda D}{d}\]
Where:
- \(\beta\) is the fringe separation (distance between consecutive bright or dark fringes).
- \(\lambda\) is the wavelength of the light.
- \(D\) is the distance from the slits to the screen.
- \(d\) is the separation between the two slits.
Calculating the Fringe Separation
Now, we can substitute the converted values into the formula:
\[\beta = \frac{(5 \times 10^{-7} \text{ m}) \times (2 \text{ m})}{1 \times 10^{-4} \text{ m}}\]
Let's perform the calculation step-by-step:
\[\beta = \frac{10 \times 10^{-7} \text{ m}^2}{1 \times 10^{-4} \text{ m}}\]
\[\beta = \frac{10^{-6} \text{ m}^2}{10^{-4} \text{ m}}\]
\[\beta = 10^{-6 - (-4)} \text{ m}\]
\[\beta = 10^{-6 + 4} \text{ m}\]
\[\beta = 10^{-2} \text{ m}\]
Converting Result to Centimeters
The result is in meters. Let's convert it to centimeters (cm) as the options are given in cm. \(1 \text{ m} = 100 \text{ cm}\).
\[\beta = 10^{-2} \text{ m} = 0.01 \text{ m}\]
\[\beta = 0.01 \times 100 \text{ cm}\]
\[\beta = 1 \text{ cm}\]
Thus, the fringe separation is 1 cm.
| Parameter | Given Value | Converted Value (SI Units) |
|---|---|---|
| Slit separation (\(d\)) | 0.1 mm | \(1 \times 10^{-4} \text{ m}\) |
| Screen distance (\(D\)) | 2 m | 2 m |
| Wavelength (\(\lambda\)) | 500 nm | \(5 \times 10^{-7} \text{ m}\) |
| Fringe separation (\(\beta\)) | ? | 1 cm |
Conclusion on Fringe Separation
Using the formula for fringe separation and the given values for slit separation, screen distance, and wavelength, we calculated the fringe separation to be 1 cm.
Revision Table: Young's Double-Slit Experiment
| Term | Symbol | Description | SI Unit |
|---|---|---|---|
| Wavelength | \(\lambda\) | Distance between successive crests or troughs of a wave. | Meter (m) |
| Slit Separation | \(d\) | Distance between the centers of the two narrow slits. | Meter (m) |
| Screen Distance | \(D\) | Distance from the plane of the slits to the observation screen. | Meter (m) |
| Fringe Separation (Fringe Width) | \(\beta\) | Distance between the centers of two consecutive bright fringes or two consecutive dark fringes on the screen. | Meter (m) |
Additional Information: Understanding Interference Fringes
Young's double-slit experiment is a fundamental demonstration of the wave nature of light. When coherent light passes through two narrow slits, the waves diffract and overlap on the screen. This overlap leads to interference.
- Constructive Interference: Occurs when crests meet crests or troughs meet troughs, resulting in bright fringes (maxima). The path difference from the two slits to the screen is an integer multiple of the wavelength (\(m\lambda\), where \(m = 0, 1, 2, ...\)).
- Destructive Interference: Occurs when crests meet troughs, resulting in dark fringes (minima). The path difference is a half-integer multiple of the wavelength (\((m + 1/2)\lambda\), where \(m = 0, 1, 2, ...\)).
The fringes are equally spaced on the screen, provided the screen distance \(D\) is much larger than the slit separation \(d\), and the angle of diffraction is small. The fringe separation \(\beta\) is directly proportional to the wavelength \(\lambda\) and the screen distance \(D\), and inversely proportional to the slit separation \(d\).
Important Questions from Ray Optics and Optical Instruments
The graph correctly representing the variation of image distance v for a convex lens of focal length f versus object distance u is:
Resolving power of a telescope can be increased by increasing:
Match List - I with List - II.
List - I List - II (A) Contracting of Eye ball (I) Myopia (B) Controls the shape of eye lens (II) Cornea (C) Elongation of eye ball (III) Ciliary Muscle (D) Control the light entering in eyes (IV) Hypermetropia Choose the correct answer from the options given below:
Four lenses of focal length ±5cm and ±200cm are available for making a telescope. To produce the largest magnification, the focal length of the eyepiece should be:
Which of the following statements are correct?
(A) When light rays undergo two internal reflections inside a raindrop, a secondary rainbow is formed.
(B) The angle between the emergent ray and the angle of the prism is called the angle of deviation.
(C) Light undergoes successive total internal reflections as it moves through an optical fiber.
(D) A telescope provides angular magnification of distant objects.
(E) A simple magnifier is a diverging lens of small focal length.
Choose the correct answer from the options given below:
